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As a part of a problem I'm working on, I think that I need to show that for any set A in any topological space,

$ \overline{(\overline{A^o})^o} = \overline{A^o}$

where the bar denotes closure and the notation $A^o$ denotes the interior of a set.

I have already convinced myself that in general, one cannot assume that $(\overline{A^o})^o = A^o$ since it does not work on the set $(1,2)\cup(2,3)$ in the reals.

But since $(\overline{A^o})^o \subset \overline{A^o}$ by basic properties of the interior, its closure must also be a subset: $ \overline{(\overline{A^o})^o} \subset \overline{A^o}$.

The inclusion in the other direction is messing with me -- it could simply be the excess of repeating symbols. I'm betting it's really simple, even just hinting at me what property to use on which set should be enough. Any suggestions?

Asaf Karagila
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Opal E
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    Does one of $A^o$ and $(\overline{A^o})^o$ contain the other? – Daniel Fischer May 06 '17 at 22:04
  • I guess that since $A^o \subset \overline{A^o}$ then the interior of both sets maintains the inclusion? That gets me there, thanks! – Opal E May 06 '17 at 22:08
  • See my note http://at.yorku.ca/p/a/c/a/24.htm for full proofs. Identity p) is what you want. But don't skip the notation part. – Henno Brandsma May 06 '17 at 22:12
  • That's the problem I'm working on -- I'll wait to check my answer there until I'm done with the second sequence of sets! Thanks, though! – Opal E May 06 '17 at 22:13
  • @DanielFischer your comment got me what I needed & is much clearer than the other answer -- if you add it in the answers, I'll be happy to accept it. – Opal E May 07 '17 at 20:22
  • Note: This is a counterpart to the claim posted at https://math.stackexchange.com/questions/2279308/is-the-interior-of-the-closure-of-the-interior-of-the-closure-of-a-set-equal-to , which says the same thing but with the roles of "closure" and "interior" swapped. (Of course, it is equivalent to that claim, because if you can replace all subsets by their complements, then closures and interiors trade places.) – darij grinberg May 23 '19 at 14:22

2 Answers2

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$A^o$ is an open subset contained in $\overline{A^o}$, so we have

$$A^o \subset \bigl(\overline{A^o}\bigr)^o.$$

Taking the closure yields $\overline{A^o} \subset \overline{\bigl(\overline{A^o}\bigr)^o}$. Together with the inclusion $\overline{\bigl(\overline{A^o}\bigr)^o} \subset \overline{A^o}$ that you got from $\bigl(\overline{A^o}\bigr)^o \subset \overline{A^o}$, this yields equality.

By a similar argument, or by taking complements and using this result, one sees that also $A \mapsto \bigl(\overline{A}\bigr)^o$ is an idempotent operator.

Daniel Fischer
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cl int cl int A = cl int A.
Proof.
cl int cl int A subset cl cl int A = cl int A
= cl int int A subset cl int cl int A.

Likewise the dual statement
int cl int cl A = int cl A
which also comes by taking the complement of both
sides of the previous result and replacing A^c with A.

William Elliot
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